MathDB

3

Part of 2013 Vietnam National Olympiad

Problems(2)

colinear and fixed point

Source: Problem 3 VMO 2013 day 1

1/11/2013
Let ABCABC be a triangle such that ABCABC isn't a isosceles triangle. (I)(I) is incircle of triangle touches BC,CA,ABBC,CA,AB at D,E,FD,E,F respectively. The line through EE perpendicular to BIBI cuts (I)(I) again at KK. The line through FF perpendicular to CICI cuts (I)(I) again at LL.JJ is midpoint of KLKL. a) Prove that D,I,JD,I,J collinear. b) B,CB,C are fixed points,AA is moved point such that ABAC=k\frac{AB}{AC}=k with kk is constant.IE,IFIE,IF cut (I)(I) again at M,NM,N respectively.MNMN cuts IB,ICIB,IC at P,QP,Q respectively. Prove that bisector perpendicular of PQPQ through a fixed point.
geometrygeometric transformationreflectionperpendicular bisectorgeometry proposed
How many ordered 6-tuples satisfy the system

Source: Vietnam National Mathematical Olympiad 2013, P7, Day 1

1/12/2013
Find all ordered 6-tuples satisfy following system of modular equation: ab+a′b′≡1ab+a'b' \equiv 1 (mod 15) bc+b′c′≡1bc+b'c' \equiv 1 (mod 15) ca+c′a′≡1ca+c'a' \equiv 1 (mod 15) Given that a,b,c,a′,b′,c′ϵ(0;1;2;...;14)a,b,c,a',b',c' \epsilon (0;1;2;...;14)
number theory proposednumber theory